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Sagot :
Given the compound inequality;
[tex]-3<-10(x+15)\le7[/tex]We would begin by simplifying the parenthesis as follows;
[tex]\begin{gathered} -3<-10(x+15) \\ \text{AND} \\ -10(x+15)\le7 \end{gathered}[/tex]We shall now solve each part one after the other;
[tex]\begin{gathered} -3<-10(x+15) \\ -3<-10x-150 \\ \text{Collect all like terms and we'll have;} \\ -3+150<-10x \\ 147<-10x \\ \text{Divide both sides by -10} \\ \frac{-147}{10}>x \end{gathered}[/tex]We can switch sides, and in that case the inequality sign would also "flip" over, as shown below;
[tex]\begin{gathered} \frac{-147}{10}>x \\ \text{Now becomes;} \\ x<\frac{-147}{10} \end{gathered}[/tex]For the other part of the compound inequality;
[tex]\begin{gathered} -10(x+15)\le7 \\ -10x-150\le7 \\ \text{Collect all like terms and we'll have;} \\ -10x\le7+150 \\ -10x\le157 \\ \text{Divide both sides by -10} \\ \frac{-10x}{-10}\le\frac{157}{-10} \\ x\ge-\frac{157}{10} \end{gathered}[/tex]Therefore, the values are;
[tex]\begin{gathered} x<-\frac{147}{10} \\ \text{And } \\ x\ge-\frac{157}{10} \\ \text{Hence;} \\ -\frac{157}{10}\le x<-\frac{147}{10} \end{gathered}[/tex]Written in interval notation, this now becomes;
[tex]\lbrack-\frac{157}{10},-\frac{147}{10})[/tex]
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